Linear Algebra and the C Language/a0bc

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00c.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ /* ------------------------------------ */ #define RA R5 #define CA C5 #define Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double xy[6] ={ 1, 10, 2, 1, 3, -10 }; double ab[RA*(CA+Cb)]={ /* x**2 y**2 x y = 0 */ +1, +0, +0, +0, +0, +1, +0, +1, +0, +0, +0, +1, +1, +100, +1, +10, +1, +0, +4, +1, +2, +1, +1, +0, +9, +100, +3, -10, +1, +0, }; double **XY = ca_A_mR(xy,i_mR(R3,C2)); double **Ab = ca_A_mR(ab,i_Abr_Ac_bc_mR(RA,CA,Cb)); double **A = c_Ab_A_mR(Ab,i_mR(RA,CA)); double **b = c_Ab_b_mR(Ab,i_mR(RA,Cb)); double **Q = i_mR(RA,CA); double **R = i_mR(CA,CA); double **invR = i_mR(CA,CA); double **Q_T = i_mR(CA,RA); double **invR_Q_T = i_mR(CA,RA); double **x = i_mR(CA,Cb); // x = invR * Q_T * b clrscrn(); printf("\n"); printf(" Find the coefficients a, b, c, d of a circle \n\n"); printf(" ax**2 + ay**2 + bx + cy + d = 0 \n\n"); printf(" that passes through these three XY. \n\n"); printf(" x y"); p_mR(XY,S5,P0,C6); printf(" Using the given points, we obtain this matrix.\n"); printf(" (a = 1. This is my choice)\n\n"); printf(" Ab :\n"); printf(" x**2 y**2 x y = 0 "); p_mR(Ab,S7,P2,C6); stop(); clrscrn(); QR_mR(A,Q,R); printf(" Q :"); p_mR(Q,S10,P4,C6); printf(" R :"); p_mR(R,S10,P4,C6); stop(); clrscrn(); transpose_mR(Q,Q_T); printf(" Q_T :"); pE_mR(Q_T,S12,P4,C6); invgj_mR(R,invR); printf(" invR :"); pE_mR(invR,S12,P4,C6); stop(); clrscrn(); printf(" Solving this system yields a unique\n" " least squares solution, namely \n\n"); mul_mR(invR,Q_T,invR_Q_T); mul_mR(invR_Q_T,b,x); printf(" x = invR * Q_T * b :"); p_mR(x,S10,P2,C6); printf(" The coefficients a, b, c, d of the curve are : \n\n" " %+.2fx**2 %+.2fy**2 %+.2fx %+.2fy %+.2f = 0\n\n" ,x[R1][C1],x[R2][C1],x[R3][C1],x[R4][C1],x[R5][C1]); stop(); f_mR(XY); f_mR(A); f_mR(b); f_mR(Ab); f_mR(Q); f_mR(Q_T); f_mR(R); f_mR(invR); f_mR(invR_Q_T); f_mR(x); return 0; } /* ------------------------------------ */ /* ------------------------------------ */ Screen output example: Find the coefficients a, b, c, d of a circle ax**2 + ay**2 + bx + cy + d = 0 that passes through these three XY. x y +1 +10 +2 +1 +3 -10 Using the given points, we obtain this matrix. (a = 1. This is my choice) Ab : x**2 y**2 x y = 0 +1.00 +0.00 +0.00 +0.00 +0.00 +1.00 +0.00 +1.00 +0.00 +0.00 +0.00 +1.00 +1.00 +100.00 +1.00 +10.00 +1.00 +0.00 +4.00 +1.00 +2.00 +1.00 +1.00 +0.00 +9.00 +100.00 +3.00 -10.00 +1.00 +0.00 Press return to continue. Q : +0.1005 -0.1023 -0.3709 +0.9175 -0.0082 +0.0000 +0.0101 -0.0043 -0.0095 -0.9999 +0.1005 +0.9068 +0.3402 +0.2277 +0.0055 +0.4020 -0.3993 +0.7909 +0.2310 -0.0096 +0.9045 +0.0881 -0.3481 -0.2299 +0.0046 R : +9.9499 +100.9058 +3.6181 -7.6383 +1.4071 -0.0000 +99.0960 +0.3725 +7.7879 +0.5956 -0.0000 +0.0000 +0.8777 +7.6742 +0.7830 +0.0000 -0.0000 +0.0000 +4.8075 +0.2288 +0.0000 +0.0000 +0.0000 +0.0000 +0.0005 Press return to continue. Q_T : +1.0050e-01 +0.0000e+00 +1.0050e-01 +4.0202e-01 +9.0453e-01 -1.0234e-01 +1.0091e-02 +9.0678e-01 -3.9927e-01 +8.8069e-02 -3.7088e-01 -4.2823e-03 +3.4023e-01 +7.9089e-01 -3.4810e-01 +9.1750e-01 -9.5115e-03 +2.2771e-01 +2.3104e-01 -2.2993e-01 -8.1722e-03 -9.9989e-01 +5.5283e-03 -9.6144e-03 +4.5668e-03 invR : +1.0050e-01 -1.0234e-01 -3.7088e-01 +9.1750e-01 -8.1722e-03 +0.0000e+00 +1.0091e-02 -4.2823e-03 -9.5115e-03 -9.9989e-01 +0.0000e+00 -0.0000e+00 +1.1393e+00 -1.8187e+00 -9.9009e+02 +0.0000e+00 -0.0000e+00 +0.0000e+00 +2.0801e-01 -9.9012e+01 -0.0000e+00 +0.0000e+00 -0.0000e+00 +0.0000e+00 +2.0802e+03 Press return to continue. Solving this system yields a unique least squares solution, namely x = invR * Q_T * b : +1.00 +1.00 +996.00 +100.00 -2097.00 The coefficients a, b, c, d of the curve are : +1.00x**2 +1.00y**2 +996.00x +100.00y -2097.00 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.00*x^2 +1.00*y^2 +996.00*x +100.00*y -2097.00; endfunction f (+1,+10) f (+2,+1) f (+3,-10)